I have an exam tomorrow and I'm seeking advice on if I can square root the variance of the mle $\theta^2$ to get te variance of $\theta$. Does the invariance property apply here?
Question:
$\theta$ is the frequency of an allele causing a mendelian recessive disease, and the probability that an individual is affected is $\theta^2$. A random sample of size n individuals is taken from a very large population, and x individuals are observed to be affected wit hthe disease.
What is the approximate distribution of the MLE when the sample size is large?
I've gotten the mle to be $\sqrt{ \frac{X}{n}}$. Hence,
The mean is:
$E[\hat{\theta}] = \sqrt{E[X]/n} = \theta$
Which is easy cause the E just goes into the square root. However, I dont think I can do this for the variance right? So my question is, can I do:
$Var[\hat{\theta^2}] = Var[\frac{X}{n}] = \frac{1}{n^2}Var[X] = \frac{p(1-p)}{n}$ where p = $\theta^2$
Then,
$Var[\hat{\theta}] = \sqrt{Var[\hat{\theta^2}]} = \sqrt{\frac{p(1-p)}{n}} $
Or do I have to find the distribution for Root of X?
Thank you!
If you're familiar with the Delta Method, then we can apply it here:
Following your lead, let $p = \theta^2$, $n$ be the sample size and $x$ the number of individuals with the disease. then we know the mle for $p$ is $\hat{p}_n=\frac{x}{n}$. As long as $p \notin \{0,1\}$ then the mle will be asymptotically normal:
$$\sqrt{n}[\hat{p}_n-p] \xrightarrow{D}\mathcal{N}(0,p(1-p))$$
Let's define $g(p) = \sqrt{p} \implies g'(p) = \frac{1}{2\sqrt{p}}$. By the Delta Method we get the asymptotic distribution of $\sqrt{\hat{p}_n}$:
$$\sqrt{n}[\sqrt{\hat{p}}_n-\sqrt{p}] \xrightarrow{D}\mathcal{N}\left(0,p(1-p)\left[\frac{1}{2\sqrt{p}}\right]^2\right) = \mathcal{N}\left(0,\frac{1-p}{4}\right)$$
Now, since $\frac{x}{n}=\hat{p} = \widehat{\theta^2}$ is a maximum likelihood estimator, then we can indeed invoke invariance of mles to say that $\hat{\theta}_n = \sqrt{\frac{x}{n}}$ is the mle of $\theta$. The asymptotic distribution of $\hat{\theta}_n$ is given by the formula above by substituting $\theta = \sqrt{p}$:
$$\sqrt{n}[\hat{\theta}_n-\theta] \xrightarrow{D} \mathcal{N}\left(0,\frac{1-\theta^2}{4}\right) \implies \hat{\theta} \;\dot{\sim}\; \mathcal{N}\left(\theta,\frac{1-\theta^2}{4n}\right)$$